Solve Using the Quadratic Formula r^2-8r-20=0

$r^{2} - 8 r - 20 = 0$

Step 1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values $a = 1$ , $b = - 8$ , and $c = - 20$ into the quadratic formula and solve for $r$ .

$\frac{8 \pm \sqrt{{(- 8)}^{2} - 4 \cdot (1 \cdot - 20)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 8$ to the power of $2$ .

$r = \frac{8 \pm \sqrt{64 - 4 \cdot 1 \cdot - 20}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 20$ .

Multiply $- 4$ by $1$ .

$r = \frac{8 \pm \sqrt{64 - 4 \cdot - 20}}{2 \cdot 1}$

Multiply $- 4$ by $- 20$ .

$r = \frac{8 \pm \sqrt{64 + 80}}{2 \cdot 1}$

Add $64$ and $80$ .

$r = \frac{8 \pm \sqrt{144}}{2 \cdot 1}$

Rewrite $144$ as $12^{2}$ .

$r = \frac{8 \pm \sqrt{12^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$r = \frac{8 \pm 12}{2 \cdot 1}$

Multiply $2$ by $1$ .

$r = \frac{8 \pm 12}{2}$

Simplify $\frac{8 \pm 12}{2}$ .

$r = 4 \pm 6$

Step 4

The final answer is the combination of both solutions.

$r = 10, - 2$